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I was watching the video from a few years ago where SN15 did the flip manoeuvre and landing. The actuated fins (lower flaps) can help Starship to pitch while during its descent. However, I am uncertain when it produces the most drag. At first I thought that the most drag is produce when the wings are horizontal since it would have the largest projected area. $$Drag = \frac{1}{2} \cdot Area \cdot Cd \cdot Air\ density$$

enter image description here

However, when the wings are bent inwards, then even though the projected area is smaller, the air pressure underneath the wings is larger.

enter image description here

This means that when Starship has it wings horizontal, it has a wider projected area, but lower air pressure underneath its wings. When its wings are bent inwards, its projected area isn't as wide, but the pressure underneath is larger. So when does Starship produce the most drag with it wings?

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  • $\begingroup$ What does exactly "direction of velocity" mean? Is the rocket or the relative wind going down in your picture? $\endgroup$
    – sophit
    Commented Mar 30 at 17:16
  • $\begingroup$ @sophit The rocket is going down. It is just the direction of the rocket $\endgroup$ Commented Mar 30 at 17:22
  • $\begingroup$ Ah ok, then the fins are actually pointing upward, the other way around. And then they close just before the touchdown in order to let the bottom of the rocket sink and rotate perpendicular to the ground $\endgroup$
    – sophit
    Commented Mar 30 at 18:27
  • $\begingroup$ @sophit Yes, that is correct. Starship does do that, but I was curious to know in which of these 2 particular positions the wings would produce the most lift. When the wings are bent upwards, the projected area is less and the pressure underneath the wings is less as well meaning that there is less lift. Of course they put it in that position because it is practical since they have better control. $\endgroup$ Commented Mar 30 at 18:45
  • $\begingroup$ imo title should mention drag rather than lift $\endgroup$
    – user21228
    Commented Mar 30 at 19:02

2 Answers 2

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Because the Starship falls horizontally you are correct to have Cdrag rather than Clift in your equation.

Based on coefficient of drag shapes here, having the fins canted down may produce more drag, but only up to a certain point.

As fin angle progresses downward, Cdrag increases but frontal Area decreases. At some point a drag maxima will occur.

But one can definitely reduce drag by canting the fins upward, giving a workable pitch control solution. One may want to avoid pitching the fins down, as this would cause roll instability in a falling object anyways.

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    $\begingroup$ It would be cool to see at what point it would have the maximum lift. I think that when the wings bend downwards there is probably a function between the loss due to less projected area and the gain due to higher air pressure underneath the wings. One would probably have to use a bit of calculus to calculate at which angle there is the maximum lift $\endgroup$ Commented Mar 30 at 13:12
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    $\begingroup$ Interesting fact: that's why the Super Heavy booster's chines are not at 180° from each other: youtu.be/3Ux6B3bvO0w?t=630 $\endgroup$ Commented Mar 30 at 14:01
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In the subsonic re-entry phase, Starship is basically falling free from the sky laying on its belly with the forward (upper) and rear (lower) fins stabilising the free fall and keeping it parallel to the ground (source):

enter image description here

I suppose it is kept in that horizontal position because it is the one offering the biggest surface to the airflow i.e. creating the biggest drag and therefore the slowest speeds.

The attitude of the fins is actually the opposite of what you've drawn i.e. they are pointing slightly upward to give the aircraft stability around the roll axis. It works just like the dihedral stability given by the wings slightly pointing upward of any jetliner.

So when does Starship produce the most drag with its wings?

To come to your question: as usual for such questions "Fluid Dynamic Drag" by "Hörner" contains the answer. In particular, figure 34 at page 3-18 shows the drag coefficient for wedges of different (half) vertex angles:

enter image description here

Even if in that plot we miss the circular section of the rocket itself, it can be anyway used to give a first plausible estimation for your question.

Angles <90° represent wedges with their tips invested by the airflow (upper wedge); at 90° we have a flat plate; while at angles >90° we have wedges with their cavity invested by the airflow (wedge in the middle-right). As can be intuitively expected, the drag coefficient increases with the vertex angle.

So, if for example:

  • the fins are pointing 30° upward, their $C_d$ would be around 1.3 (we look at the "2-dimensional" plot);
  • at 90° it would be 2 (flat plate);
  • and at 30° downward (more or less like in your picture) it would be around 2.1 (90+30=120°).

All of these coefficients have been obtained adimensionalising in respect to the projected area: said 1 the area at 90°, its projection becomes 0.866 at 30°. That means that if at 90° we have an aerodynamic drag proportional to $D\propto A\times C_d = 1\times 2=2$, then at 30° upward it is $0.866\times 1.3=1.126$ while at 30° downward it becomes $0.866\times 2.1=1.82$.

So, the biggest drag should be obtained at 90°.

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